How many grams of C would be necessary to produce 790. grams of CaCO3 by the sequence of reactions shown below?

Step 1: CaO + 3C → CaC2 + CO

Step 2: CaC2 + N2 → CaCN2 + C

Step 3: CaCN2 + 3H2O → CaCO3 + 2NH3

In the end for the carbon

3 C --> CO + C + CaCO3
so in the end we need 3 C atoms for every molecule of CaCO3

C = 12 g
Ca = 40 g
3O = 3*16 = 48 g
so CaCO3 = 100 g/mol
so we get 7.90 mols of CaCO3
for that we need 3*7.9 = 23.7 mols of C
23.7 * 12 = 284.4 grams of Carbon

we got carbon as a result of step 2 so we could put it back in to step 1

then we would only need two carbons fo each CaCO3

2 * 7.9 = 15.8 mols C

15.8 * 12 g/mol = 189.6 grams of carbon

To determine the number of grams of C required to produce 790 grams of CaCO3, we need to calculate the stoichiometry for each step in the sequence of reactions.

Let's start by finding the molar mass of CaCO3. The molar mass of Ca is 40.08 g/mol, the molar mass of C is 12.01 g/mol, and the molar mass of O is 16.00 g/mol.

CaCO3 = (40.08 g/mol Ca) + (12.01 g/mol C) + (3 * 16.00 g/mol O)
CaCO3 = 40.08 g/mol + 12.01 g/mol + 48.00 g/mol
CaCO3 = 100.09 g/mol

Now, let's calculate the molar mass of Step 3. It shows that 1 mol of CaCN2 reacts with 3 mol of H2O to produce 1 mol of CaCO3 and 2 mol of NH3.

Next, we'll use the molar mass of CaCO3 to find the number of moles of CaCO3 produced. We need to convert the given mass of CaCO3 (790 g) to moles.

Moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3
Moles of CaCO3 = 790 g / 100.09 g/mol
Moles of CaCO3 = 7.89 mol

Since 1 mol of CaCO3 is produced from 1 mol of CaCN2, we can say that 7.89 mol of CaCN2 are required.

Now, using the stoichiometry of Step 2, we know that 1 mol of CaCN2 is produced from 1 mol of C. Therefore, we need 7.89 mol of C.

Finally, we can calculate the mass of C required using the molar mass of C:

Mass of C = Moles of C * Molar mass of C
Mass of C = 7.89 mol * 12.01 g/mol
Mass of C = 94.59 g

Therefore, approximately 94.59 grams of C would be necessary to produce 790 grams of CaCO3.